A-LevelPhysicsElectricity9702/22 · Oct/Nov 2017 · Q612 Marks

State what is meant by an electric current.

(a)State what is meant by an electric current.[1]
(b)(i)A metal wire has length L and cross-sectional area A, as shown in Fig. 6.1. [Figure 6.1] I is the current in the wire, n is the number of free electrons per unit volume in the wire, v is the average drift speed of a free electron and e is the charge on an electron. State, in terms of A, e, L and n, an expression for the total charge of the free electrons in the wire.[1]
(b)(ii)Use your answer in (i) to show that the current I is given by the equation I = nAve.[2]
(c)(i)A metal wire in a circuit is damaged. The resistivity of the metal is unchanged but the cross-sectional area of the wire is reduced over a length of 3.0mm, as shown in Fig. 6.2. [Figure 6.2] The wire has diameter d at cross-section X and diameter 0.69d at cross-section Y. The current in the wire is 0.50A. Determine the ratio average drift speed of free electrons at cross-section Y / average drift speed of free electrons at cross-section X[2]
(c)(ii) 1.The main part of the wire with cross-section X has a resistance per unit length of 1.7 × 10⁻² Ωm⁻¹. For the damaged length of the wire, calculate the resistance per unit length,[2]
(c)(ii) 2.the power dissipated.[2]
(c)(iii)The diameter of the damaged length of the wire is further decreased. Assume that the current in the wire remains constant. State and explain qualitatively the change, if any, to the power dissipated in the damaged length of the wire.[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

[B1] flow of charge carriers
[B1] nALe
[B1] (t is time taken for electrons to move length L) I = Q/t
[B1] I = nALe/t or I = nALe/ (L/v) or I = nAvte/t and I = nAve
[C1] ratio = area at X/area at Y = [πd²/4]/[π(0.69d)²/4] or d²/(0.69d)² or 1/0.69²
[A1] = 2.1
[C1] R = ρL/A or R/L ∝ 1/A
[A1] resistance per unit length = 1.7 × 10⁻² × (area at X/area at Y) = 1.7 × 10⁻² × 2.1 = 3.6 × 10⁻² Ωm⁻¹
[C1] P = I²R or P = V²/R
[A1] R = 3.6 × 10⁻² × 3.0 × 10⁻³ (= 1.08 × 10⁻⁴ Ω) AND P = 0.50² × 1.08 × 10⁻⁴ = 2.7 × 10⁻⁵ W
[M1] (cross-sectional area decreases so) resistance increases
[A1] (P = I²R, so) power increases

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